Dynamics Chapter 15: Matlab Solutions

15-1. A man kicks the 150-g ball such that it leaves the ground at an angle of 60° and strikes the ground at the same elevation a distance of 12 m away. Determine the impulse of his foot on the ball at A. Neglect the impulse caused by the ball's weight while it's being kicked. Matlab % Given m = 0.15; % Mass of ball (kg) theta = 60 * pi/180; % Angle of projection (rad) s = 12; % Horizontal range (m) % Find initial velocity components vy0 = input('Enter the initial vertical velocity (m/s): '); vx0 = vy0 * tan(theta); % Calculate time of flight t = (vx0 + sqrt(vx0^2 + 2*9.81*s)) / 9.81; % Calculate impulse I = m * (vx0 * cos(theta) + vy0 * sin(theta)); disp(['The impulse of his foot on the ball is ', num2str(I), ' N.s']); 15-2. A 20-lb block slides down a 30° inclined plane with an initial velocity of 2 ft/s. Determine the velocity of the block in 3 s if the coefficient of kinetic friction between the block and the plane is mk = 0.25. Matlab % Given m = 20 / 32.2; % Mass of block (slugs) theta = 30 * pi/180; % Angle of inclined plane (rad) v0 = 2; % Initial velocity (ft/s) mu_k = 0.25; % Coefficient of kinetic friction t = 3; % Time (s) % Calculate acceleration ax = 9.81 * (sin(theta) - mu_k * cos(theta)); % Acceleration (ft/s^2) % Calculate final velocity vx = v0 + ax * t; % Final velocity (ft/s) disp(['The velocity of the block after 3 seconds is ', num2str(vx), ' ft/s']); 15-3. The uniform beam has a weight of 5000 lb. Determine the average tension in each of the two cables AB and AC if the beam is given an upward speed of 8 ft/s in 1.5 s starting from rest. Neglect the mass of the cables. Matlab % Given m = 5000 / 32.2; % Mass of beam (slugs) v = 8; % Final upward speed (ft/s) t = 1.5; % Time to achieve final speed (s) % Calculate impulse I = m * v; % Impulse (lbf.s) % Calculate tension force F = I / (2 * t); % Tension force (lbf) disp(['The average tension in each cable is ', num2str(F), ' lbf']); 15-4. Each of the cables can sustain a maximum tension of 5000 lb. If the uniform beam has a weight of 5000 lb, determine the shortest time possible to lift the beam with a speed of 10 ft/s starting from rest. Matlab % Given m = 5000 / 32.2; % Mass of beam (slugs) v = 10; % Final upward speed (ft/s) F_max = 5000; % Maximum tension force (lbf) W = 5000; % Weight of beam (lbf) % Calculate maximum possible acceleration a_max = (2 * F_max - W) / m; % Maximum acceleration (ft/s^2) % Calculate minimum time t_min = v / a_max; % Minimum time (s) disp(['The shortest time possible to lift the beam is ', num2str(t_min), ' seconds']); 15-5. A hockey puck is traveling to the left with a velocity of v1 = 10 m/s when it is struck by a hockey stick and given a velocity of v2 = 20 m/s as shown. Determine the magnitude of the net impulse exerted by the hockey stick on the puck. The puck has a mass of 0.2 kg. Matlab % Given m = 0.2; % Mass of puck (kg) v1 = 10; % Initial velocity (m/s) v2 = 20; % Final velocity (m/s) theta = 40 * pi/180; % Angle of final velocity (rad) % Calculate change in momentum delta_p = m * (v2 * cos(theta) - (-v1)); % Change in momentum (kg.m/s) % Calculate impulse I = abs(delta_p); % Impulse (N.s) disp(['The magnitude of the net impulse is ', num2str(I), ' N.s']); 15-6. A train consists of a 50-Mg engine and three cars, each having a mass of 30 Mg. If it takes 80 s for the train to increase its speed uniformly to 40 km/h, starting from rest, determine the force T developed at the coupling between the engine E and the first car A. The wheels of the engine provide a resultant frictional tractive force F which gives the train forward motion, whereas the car wheels roll freely. Also, determine F acting on the engine wheels. Matlab % Given m_engine = 50 * 1000; % Mass of engine (kg) m_car = 30 * 1000; % Mass of each car (kg) n_cars = 3; % Number of cars t = 80; % Time to achieve final speed (s) v_final = 40 * 1000 / 3600; % Final speed (m/s) % Calculate total mass m_total = m_engine + n_cars * m_car; % Total mass (kg) % Calculate acceleration a = v_final / t; % Acceleration (m/s^2) % Calculate tractive force on engine wheels F = m_total * a; % Tractive force (N) % Calculate force at coupling T = (m_car * n_cars) * a; % Force at coupling (N) disp(['The force developed at the coupling is ', num2str(T), ' N']); disp(['The tractive force on the engine wheels is ', num2str(F), ' N']); 15-7. Crates A and B weigh 100 lb and 50 lb, respectively. If they start from rest, determine their speed when t = 5 s. Also, find the force exerted by crate A on crate B during the motion. The coefficient of kinetic friction between the crates and the ground is mk = 0.25. Matlab % Given m_A = 100 / 32.2; % Mass of crate A (slugs) m_B = 50 / 32.2; % Mass of crate B (slugs) mu_k = 0.25; % Coefficient of kinetic friction t = 5; % Time (s) % Calculate normal forces N_A = m_A * 32.2; % Normal force on crate A (lbf) N_B = m_B * 32.2; % Normal force on crate B (lbf) % Calculate frictional forces f_A = mu_k * N_A; % Frictional force on crate A (lbf) f_B = mu_k * N_B; % Frictional force on crate B (lbf) % Calculate net forces F_A = N_A - f_A; % Net force on crate A (lbf) F_B = N_B - f_B; % Net force on crate B (lbf) % Calculate acceleration a = (F_A - F_B) / (m_A + m_B); % Acceleration (ft/s^2) % Calculate final velocity v = a * t; % Final velocity (ft/s) % Calculate force exerted by crate A on crate B F_AB = m_A * a; % Force exerted by crate A on crate B (lbf) disp(['The speed when t = 5 s is ', num2str(v), ' ft/s']); disp(['The force exerted by crate A on crate B is ', num2str(F_AB), ' lbf']); 15-8. The automobile has a weight of 2700 lb and is traveling forward at 4 ft/s when it crashes into the wall. If the impact occurs in 0.06 s, determine the average impulsive force acting on the car. Assume the brakes are not applied. If the coefficient of kinetic friction between the wheels and the pavement is mk = 0.3, calculate the impulsive force on the wall if the brakes were applied during the crash. The brakes are applied to all four wheels so that all the wheels slip. Matlab % Given m = 2700 / 32.2; % Mass of car (slugs) v0 = 4; % Initial velocity (ft/s) t = 0.06; % Impact time (s) mu_k = 0.3; % Coefficient of kinetic friction % Case 1: Brakes not applied % Calculate impulse I = m * v0; % Impulse (lbf.s) % Calculate average impulsive force F_avg = I / t; % Average impulsive force (lbf) disp(['Case 1: The average impulsive force is ', num2str(F_avg), ' lbf']); % Case 2: Brakes applied % Calculate normal force N = m * 32.2; % Normal force (lbf) % Calculate frictional force f = mu_k * N; % Frictional force (lbf) % Calculate impulse on wall I_wall = I - f * t; % Impulse on wall (lbf.s) % Calculate impulsive force on wall F_wall = I_wall / t; % Impulsive force on wall (lbf) disp(['Case 2: The impulsive force on the wall is ', num2str(F_wall), ' lbf']); 15-9. The 200-kg crate rests on the ground for which the coefficients of static and kinetic friction are ms = 0.5 and mk = 0.4, respectively. The winch delivers a horizontal towing force T to its cable at A which varies as shown in the graph. Determine the speed of the crate when t = 4 s. Originally the tension in the cable is zero. Hint: First determine the force needed to begin moving the crate. Matlab % Given m = 200; % Mass of crate (kg) mu_s = 0.5; % Coefficient of static friction mu_k = 0.4; % Coefficient of kinetic friction t_start = sqrt((mu_s * m * 9.81) / (400 / 2)); % Time to start moving (s) % Define force function T = @(t) 400 * sqrt(t / 2); % Calculate net force if t <= t_start F_net = T(t) - mu_s * m * 9.81; else F_net = T(t) - mu_k * m * 9.81; end % Calculate acceleration a = F_net / m; % Acceleration (m/s^2) % Calculate velocity v = integral(@(t) a, 0, t); % Velocity (m/s) % Display result disp(['The speed of the crate when t = 4 s is ', num2str(v(4)), ' m/s']); 15-10. The 50-kg crate is pulled by the constant force P. If the crate starts from rest and achieves a speed of 10 m/s in 5 s, determine the magnitude of P. The coefficient of kinetic friction between the crate and the ground is mk = 0.2. Matlab % Given m = 50; % Mass of crate (kg) v_final = 10; % Final velocity (m/s) t = 5; % Time (s) mu_k = 0.2; % Coefficient of kinetic friction % Calculate acceleration a = v_final / t; % Acceleration (m/s^2) % Calculate normal force N = m * 9.81; % Normal force (N) % Calculate frictional force f = mu_k * N; % Frictional force (N) % Calculate pulling force P = m * a + f; % Pulling force (N) disp(['The magnitude of the pulling force P is ', num2str(P), ' N']);